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I have a second order Markov chain with 4 states {A,T,C,G} (the 4 DNA nucleotides).

the transition matrix looks like this:

    A    T    C    G
AA[0.1, 0.6, 0.2, 0.1]
AT[0.3, 0.1, 0.5, 0.1]
AC[0.5, 0.3,  0,  0.2]
AG[..., ..., ..., ...]
TA[..., ..., ..., ...]
TT[..., ..., ..., ...]
TC[..., ..., ..., ...]
TG[..., ..., ..., ...]
CA[..., ..., ..., ...]
CT[..., ..., ..., ...]
CG[..., ..., ..., ...]
GA[..., ..., ..., ...]
GT[..., ..., ..., ...]
GC[..., ..., ..., ...]
GG[..., ..., ..., ...]

I wanted to calculate the stationary probability vector for the 4 states to which this matrix converges. The Markov chain is regular.

In case of first order Markov chains this is easily done by calculating the limit of $P^n$ with $n\rightarrow \infty$.

I do not know how to approach the problem in case of second order Markov chains.

Also, having a limited dataset from which to determine the transition matrix, can I consider the stationary distribution of the 4 nucleotides as being the theoretical distribution I would have if I had a much larger pool from which to draw (with the same transition matrix)?

In other words, can I consider the stationary distribution like an estimation of the theoretical nucleotide frequency given the transition matrix obtained from limited data?

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  • $\begingroup$ Crossposted. (Please don't do this in the future. Choose the single best site for your question.) $\endgroup$ – cardinal May 6 '13 at 12:27
  • $\begingroup$ sorry, i didn't know this was discouraged. $\endgroup$ – Tito Candelli May 6 '13 at 12:46
  • $\begingroup$ Could you please explain the sense in which this is a "transition matrix"? Would this mean, for instance, that the chance of a transition from AC to CT is given by the value $0.3$ in the $(3,1)$ position? If that is the case, then it seems you are merely using a shorthand to describe a transition matrix on $16$ (not $4$) states, {AA, AT, ..., GG} and that you could immediately write down the $16$ by $16$ matrix $\mathbb{P}$ and solve your problem in the usual way. $\endgroup$ – whuber May 6 '13 at 14:19

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